J. oflnequal. & Appl., 1999, Vol. 3, pp. 267-277
Reprints available directly from the publisher
Photocopying permitted by license only
(C) 1999 OPA (Overseas Publishers Association) N.V.
Published by license under
the Gordon and Breach Science
Publishers imprint.
Printed in Malaysia.
On the Fundamental Polynomials for
Hermite-Fejdr Interpolation of
LagrangeType on the
Chebyshev Nodes
SIMON J. SMITH*
Division of Mathematics, La Trobe University, RO. Box 199,
Bendigo, Victoria 3552, Australia
(Received 30 June 1997," Revised 9 July 1998)
For a fixed integer rn > 0 and < k < n, let Ak,2m,n(T 37) denote the kth fundamental
polynomial for (0,
2m) Hermite-Fejdr interpolation on the Chebyshev nodes
{Xj,n=COS[(2j 1)Tr/(2n)]: < j < n}. (So Ak,2m,n(T 37) is the unique polynomial of degree
which satisfies Ak,2m,n (T, Xj,n) Skj, and whose first 2m derivatives
at most (2m + 1)n
vanish at each Xj, n.) In this paper it is established that
[Ztk,2m,n(T,x)[ ./ll,2m,n(Z 1),
k
n,
<_ x <_ 1.
It is also shown that A 1,2m,n(Z, 1) is an increasing function of n, and the best possible bound
Cm so that IAk,2m,,(T, x)l < Cm for all k, n and x E [- 1, 1] is obtained. The results generalise
those for Lagrange interpolation, obtained by P. Erdbs and G. Griinwald in 1938.
Keywords." Hermite-Fejdr interpolation; Lagrange interpolation;
Polynomial interpolation; Chebyshev nodes
AMS 1991 Subject Classification." 41A05, 41A10
* E-mail: S.Smith@bendigo.latrobe.edu.au.
Web: http://www.bendigo.latrobe.edu.au/mte/maths/staff/smith/.
267
S.J. SMITH
268
1 INTRODUCTION
_
Suppose f is a continuous real-valued function defined on the interval
[- 1, 1], and let
X= {xk,n" k- 1,2,...,n; n- 1,2,3,...}
be an infinite triangular matrix such that for all n,
> X l,n > X2,n >’’" > Xn,n
--1.
Then for each integer m>_0 there exists a unique polynomial
nm,n()[ f, x) of degree at most (m + 1)n-1 which satisfies
(r) (X,
nrn,n
f Xk,n) 50,rf(Xk,n),
< k < n, 0 < r < m.
is known as as the (0, 1,...,m) Hermite-Fej6r (HF)
interpolation polynomial off(x), and it can be expressed as
nm,n(X,f,x)
n
nm,n(, ; x) f(Xk,n)Ak,m,n(, X),
k=l
where Ak,m,n(X,x) is the unique polynomial of degree at most
(m + 1)n- such that
(r)
kmn (X, Xj,n)
< k, j < n, 0 < r < m.
O,r k,
The Ak,m,n(X X) are referred to as the fundamental polynomials for
(0, 1,..., m) HF interpolation on X, and the quantities
,m,n(X, X)
-[Ag,m,,(X, x)
k=l
and
Am n(X)
max /m n( X’, X),
-l<x<l
which are known as the Lebesgue function and Lebesgue constant,
respectively, for (0, 1,..., m) HF interpolation on X, play a crucial role
in determining the convergence behaviour of Hm,n(X,f, x) to f(x) as
HERMITE-FEJIR INTERPOLATION
269
If m 0, we obtain the familiar Lagrange interpolation process. Here
it is known (cf. [10, Section 1.3]) that there is a positive constant c so that
2
A0,n(X) > -logn + c
(1)
for any X. This leads to the classic result [5] that for any X there exists
fE C[- 1, 1] so that Ho,n(X, f, x) does not converge uniformly to fix) on
[-1, 1] as n. On the other hand, if T denotes the matrix of
Chebyshev nodes
T= cos
2n
7r
k--1,2,...,n;n=l,2,3,...
then
A0,n(T)_<-21ogn+l,
n=1,2,3,...
(See [10, Theorem 1.2].) Further, Ho,n(X,f, x) converges uniformly to
f(x) on [-1, 1] iffsatisfies the relatively mild Dini-Lipschitz condition
(See, for example, [9, Section 4.1].) Here
f(1/n)logn 0 as n
the
modulus
of
continuity
off, defined by
ofdenotes
.
coy(6)
max{If(s -f(t)l: (s,t} c [-1, 1],
Is-
_< 6}.
Thus, in terms of the magnitude of Lebesgue constants, and convergence properties of the interpolation polynomials, the Chebyshev nodes
are close to optimal for Lagrange interpolation.
The fundamental polynomials Ak,o,(T,x) and Lebesgue function
A0,,(T,x) have been studied extensively. For example, Erd6s and
Grinwald [4] obtained the following result.
THEOREM
For n-- 1, 2, 3,...,
max
max
l<k<n -l<x<l
IAko,n(T,x)l
.
71-
cot4--
or k n and
and x
The maximum is attained if and only if k
1. Furthermore, since the right-hand side is monotonic increasing to
x
S.J. SMITH
270
4/7r as n
o, it follows that for all k and n,
4
and the constant on the right-hand side is best possible.
With regard to the Lebesgue constant, Ehlich and Zeller [3]
demonstrated that for n 1,2, 3,...,
_
(2)
A0,n(T)-- A0,n(T, +1).
A proof of this result is also developed in Rivlin [10, Section 1.3], and
closely related results are presented in Brutman [1] and Gfinttner [6].
For higher-order HF interpolation, there are many similarities
between the Lagrange and (0, 1,..., m) HF processes for even values
of m. For instance, Szabados [12] extended (1) by showing that there are
positive constants Cm so that for any X,
A2m,n(X)
Cm logn,
n
1,2, 3,...
Thus, for any node system X, there exists fEC[-1,1] so that
Hzm,n(X, f, x) does not converge uniformly to f(x) on [- 1, 1] as n ec.
With regard to Lebesgue constants for the Chebyshev nodes, Byrne
et al. [2] generalized (2) by showing that
A2m,n(T)
2
/2m,n(T, +l)
rr
(2m)’ logn+O(1)
22m (m!) 2
as n
--+
oo.
The aim of this paper is to study the fundamental polynomials for
(0, 1,..., 2m) HF interpolation on the Chebyshev nodes. We obtain the
following generalization of Theorem 1.
THEOREM 2
If m >_ 0 is fixed, and n
max max
l<k<n -l<x<l
1,2, 3,..., then
[Ak,2m,n(T,x)[
1. Furthermore,
and x
or k n and x
is attained if and only ifk
IA1,2m,n(T, 1)[ is an increasing function of n, andfor all k and n,
m()2m+l-2r
[Ak,2m,n(T,x)l <. 2 Z ar,m
r---O
< x < 1,
(3)
HERMITE-FEJIR INTERPOLATION
271
where the ar,m are the coefficients & the Laurent expansion
sin2m+lz z2m+
ar,m Z2r, 0 < [Z[ < 7r.
(4)
r=0
The constant on the right-hand side of(3) is best possible.
Thus, for example, for all k, n and x
IA,2m,n(T, x)
<
[-1, 1],
4/7r
2/7r + 16/7r
3/(27r) + 40/(37r3) + 64/7r
if rn
if rn
if rn
0,
1,
2,
and the constants on the right-hand side are best possible.
Note that the corresponding problem to that considered in Theorem 2
for odd-order HF interpolation on the Chebyshev nodes is answered
by a result of Smith [11] which states that the Ak,Zm + 1,n(T, X) are nonfor all x,
negative for --1 _<x <_ 1. Thus, since
Ak,2m+l,n(T,x)
it follows that
=
max
max
l<k<n -l<x<l
]Ak2m+l,n(Z,x)[--1
and for each k the maximum is attained if and only if x=
cos[(2k 1)7r/(2n)].
The proof of Theorem 2 will be presented in Section 2. For the proof
of Theorem 1, Erd6s and Grtinwald used Riesz’s lemma [8], which
provides a lower bound for the separation of the maximum point for
the absolute value of a trigonometric polynomial and the zeros of the
polynomial, and explicit formulas for the fundamental polynomials
Alc, l,n(T, x) for (0, 1) HF interpolation on T. Because the formulas for
the Ak,m,n(T x) become increasingly more complicated with increasing
m, our method for proving Theorem 2 relies as much as possible on
zero-counting techniques and adaptations of Riesz’s method to trigonometric polynomials with multiple roots. Specific formulas for the
fundamental polynomials are used only for the final part of the proof.
2 PROOF OF THEOREM 2
For fixed m and < k < n, define the cosine polynomial tk,2m,n(O) by
tk,2m,n(O)
Ak,2m,n(Z cos 0),
S.J. SMITH
272
and for given j put
Oj
Oj,n
2j--1)
.
2"--
Then tk,2m,n(O) is the unique trigonometric polynomial of degree at
most (2m + 1)n- which satisfies
t(km,n(Oj)
O,r kd,
< k, j < n, 0 < r < 2m.
(5)
In the following sequence of lemmas the problem of finding
max0< 0< [tk,2m,n(O)[ is studied. Since this problem is equivalent to that
of finding max_l <x<llAk,2m,n(Z,x)l, the lemmas provide a proof of
Theorem 2.
LEMMA For <_ k < n, tk,2m,n (0) has zeros of order 2m at 01, 02,
On,
and of order at 0 and 7r. If k- 1, tk,2m,n’
(0) also has a single zero in each
interval (Oj, Oj+ 1)for 2 <_j <_ n-l; ifk n, then tk,2m,n(O has a single zero
in each interval (0, O + 1) for < j < n-2;/f 2 < k < n- 1, then tk,2m,nt
(0)
has a single zero in each interval (0, Oj + 1) for .<_j < n- 1, j k 1, k,
and has a zero in (Ok-l, Ok + 1) that is additional to the 2m zeros at Ok. In
all cases, tk,2m,n(O) has no other zeros in [0, 7r]. Further, tk,2m,n(O) has no
zeros in [0, 7r] apart from those given by (5), and changes sign at each of its
zeros. (Hence tk,2m,n(O) > 0 on (0k-l,0k + 1).)
-
Proof Suppose 2 < k < n- 1. By (5), tk,2m,n(O
has zeros of order 2m
at 01,02,...,0n in (0,70, and (by Rolle’s theorem) has a zero in
each interval (0, 0j+ 1) for < j_< n- 1, j k- 1, k. Further, since
tk,2m,n(Ok-1) tk,2m,n(Ok+ 1), tk,2m,n(O) has an odd number of zeros in
(0k-l,0k + 1), and so there is at least one zero of tk,2m,n(O) in (0k-l, 0k + 1)
in addition to those already identified. We have thus located
2mn+n 2zeros of tk,2m,n(O in (0, 70 Since tk,2m,n(O is odd, it also
has zeros at 0 and 7r. Hence, because tk,Zm,nt (0) has degree at most
(2m + 1)n-1, we have identified all zeros of t,Zm,n(O) in [0, 7r]. The
remaining parts of the lemma now follow immediately. The cases k
and k n are handled in a similar (indeed, slightly simpler) fashion.
LEMMA 2 For <_k <_n, the maximum of Itk,:Zm,n(O)] on the interval
0 <_ 0 <_ 7r is achieved at a unique point k k,, where f51 O,
7r, and
fhk E (Ok-l, Ok+ 1)for 2 <_ k <_ n-1. Further, tk,2m,n(qk) > O.
HERMITE-FEJIR INTERPOLATION
Proof
273
Fix k so that 2 _< k _< n- 1, and let a E [0, 7r] be such that
max Itk,2m,n(O)[
0<0<Tr
M-"
[tk,2m,n(Ol,)l.
Suppose a E (0r, 0r + 1) for some r so that 0 _< r _< n and r -J= k- 1, k. (Note
that if r 0 or n, then a 0 or 7r, respectively.) If sgn (tk,2m,n(O)) e for
0r < 0 < 0r + 1, consider
tk,2m,n(O)
f(O)
(-- 1)reMcos2m+l nO.
Now,f’(0) has zeros of order 2m at 01, 02,... On, and by Rolle’s theorem
it has a zero in each interval (0j, 0j+ 1) for _<j _< n-1 and j k-1,k.
Also,
tk,2m,n(Ok-1/2) (--1)r+k-leM,
tk,2m,n(Ok+l/2) / (-- 1)r+k-leM,
sof(Ok-1/2)f(Ok + 1/2) 0. Thusfhas a zero in [0k-1/2, 0k + 1/2], and hence
f(Ok-1/2)
f(Ok+l/2)
(by Rolle’s theorem),f’ has at least 2 zeros in (0k-l, 0k + 1) in addition to
those already identified at 0k. So we have located 2mn / n- zeros of
f’(O) in [01,
-
Note that if r 0, then f(0)= 0, and sol’(0) has an additional zero in
(0, 01). Similarly, if r= n, then f’(O) has an additional zero in (On, rr).
Finally, if < r < n- and r k- 1, k, then
f(a)
f(Or+l/2)
M- eM[ cos2m+l na[,
el&,2m,n(Or+l/2)l eM.
Iff(0r + 1/2) 0, thenf’(0) has 2 zeros in (0r, 0r + 1), while iff(Or + 1/2) 0,
thenfhas a zero between a and 0r + 1/2, and so againf’(0) has 2 zeros in
(Or, Or+l). In either case we have found an additional zero off’(0) in
(Or, Or + 1) to that already located. Overall, then, for all choices of r with
0 _< r _< n and r k- 1, k, we have located 2mn + n zeros off’(0) in (0, rr),
and since f’(O) is odd (so has zeros at 0, rr), we have identified 4mn +
2n + 2 zeros off’(0) in (-rr, rr]. This provides a contradiction, sincef’(0)
has degree 2mn + n, and so the assumption that a (Or, Or + 1) for some
r such that 0 <_ r < n and r-J: k- 1, k is incorrect. Therefore a
(0k-l,0k+ 1). Further, since tk,2m,n(O) has only one turning point in
(0k-l, 0k + 1), a is unique.
S.J. SMITH
274
The cases k
and k n are resolved in a similar manner, and so
the proof of Lemma 2 is complete.
Note that, by symmetry, tn,Zm,n(O)---tl, Zm, n(Tr--O) for all 0. Thus the
following lemma completes the proof of the first part of Theorem 2.
LEMMA 3 For 2 < k < n-1, tl,Zm,n(O > tk,Zm,n()k).
Proof For fixed k so that 2 < k < n-1, consider
g(O) tl,zm,n(O) tk,Zm,n(O %- Ok-l
which is a trigonometric polynomial of degree at most (Zm + 1)n- 1.
Also, let I denote the interval [0-zk + 1, 02n-k + 1) of width 27r. Note that
g(O) satisfies g(Oo)- 1, g(O-zk + 2)-- 1, and g(Oj)= 0 for -Zk + _<j _<
2n 2k + and j 7 0, -Zk + 2.
In I, g’(O) has zeros of order 2m at 0j for -Zk + <_ j <_ 2n 2k, and
by Rolle’s theorem g’(O) has a zero in each interval (0,0+) for
-Zk + 3 <_j <_ 2n-Zk, j
1,0. So we have identified (4m + 2)n-4
zeros ofg’(0) in I. Because g(O_k + 1) g(O-zk + 3), g’ has an odd number
of zeros in (0_k + 1, 0-2k + 3), and so g’ has a zero in (0_k + 1, 0_zk + 3) in
addition to the 2m zeros at 0_k + that have already been mentioned.
Similarly, gl has an additional zero in (0_ 1, 0), and so all (4m + 2)n 2
zeros of g(O) in I have been located.
By the above discussion, the only zeros of g(O) on (0_1, 01) are a zero
of order 2m + at 00 or else a zero of order 2m at 00 and a zero of order
at another point in the interval. In either case, g has only one turning
point on (0_1, 0), and since g(O_l)= g(O)=0 and g(Oo)> 0, it follows
that g(0)>0 if 0_1 <0<01. In particular, this gives tl,zm,n(O)>
tk,Zm,n(O %- Ok-1/Z) for 00 < 0 < 01. Since the maximum value of tl,zm,n(O)
in [00, 01] is tl,Zm,n(O), we obtain
tl,Zm,n(O) > tk,Zm,n(O), Ok-1
0
(Strict inequality holds at 0k because tl,zm,n(O) >
A similar zero-counting argument applied to
h(O)
Ok.
tk,Zm,n(Ok).)
tl,zm,n(O) tk,Zm,n(O %- Ok+l
on the interval [O_zk,OZn_Zk) shows that
tl,Zm,n(O) > tk,Zm,n(O), O < 0 < 0+.
HERMITE-FEJIR INTERPOLATION
Thus tl,2m,n(O) > tk,2m,n(O) for Ok-X <_ 0<_ Ok+X, and since
(Ok-l,0k + 1), then tl,2m,n(O) > tk,2m,n(qk).
275
bk E
The following lemma concludes the proof of Theorem 2 by establishing the monotone increasing property and limiting value of tl,2m,n(O)
as a function of n.
LEMMA 4 For fixed integer m > O, tl,2m,n(O)
is an increasing function
of n, and
lim tl,2m n(O)
2
no
ar,m
Zm()2m+l-2r
(6)
r=0
where the ar,m are given by (4).
Proof By [7, Theorem 1.1], the function
T2m,n(O)
sin 2m+l nO
2n2m+l
r--0
0
ar,m n 2r d2m-2r
2m-2"’------cOt
(2m 2r) d0
(7)
is the unique trigonometric polynomial of the form
(2m+ 1)n
T2m,n (0)
Z
(2m+ )n-
bk cos kO +
Z
ek sin kO
(8)
k=l
k=0
such that
2r)m,n()
(0,rt0d
’
0<j<2n--1_ 0<r<2m._
Now, from (7) it follows that T2m,n(O is even, and so
S2m,n(O)
Z2m,n(O2t--n) f- Z2m,n(O -)
(9)
is a cosine polynomial of degree no greater than (2m + 1)n which
satisfies
(r)
tO,r tlj,
S2m,n(Oj)
<_ j <_ n, 0 <_ r <_ 2m.
Furthermore, by (8) and (9), the cos(2m + 1)n0 term in Szm,n(O) vanishes,
and so S2m,n(O) is of degree (2m + 1)n- 1. By uniqueness considerations
S.J. SMITH
276
it follows that
tl,2m,n(O)
S2m,n(O) tl,2m,n(O) for all 0, and hence from (7),
e T2m,n
-n
n 2m+l
r=0
(2m 2r)!
LdO2m-2r
cot
O=Tr/(2n)
Now, from the well-known Laurent series for cot0 about 0, we obtain
o
cotO 20 ’=
02j_
2[’BzYl (2j)i
where the By are Bernoulli numbers. Consequently, we can write
d2m-2r
dO2m_2 cot
]
Cj,r
2(2m 2r)!
O=/(2n)
j=l
n2j-1
where the cj, are all positive, and so
m
tl,2m,n(O)
2
Z
ar,m
r=0
()
2m+l-2r
m
o
Z Z (2mar,m 2r)’ n
Cj,
2j+2m-2r"
(10)
r=0 j=l
By the definition (4) of the at,m, and the expansion
sin 0
+
=
(22,,_ 2) 02J_ o< tot
it follows that the ar,m are all positive. Thus, by (10), tl,2m,n(O) is an
increasing function of n, with limit as given by the right-hand side of (6).
Acknowledgement
The author wishes to thank the members of the La Trobe University,
Bendigo, Approximation Theory Seminar, from whom he first became
aware of the problem discussed in this paper.
References
[1] L. Brutman, On the Lebesgue function for polynomial interpolation, SIAM J.
Numer. Anal. 15 (1978), 694-704.
HERMITE-FEJIR INTERPOLATION
277
[2] G.J. Byrne, T.M. Mills and S.J. Smith, The Lebesgue constant for higher order
Hermite-Fej6r interpolation on the Chebyshev nodes, J. Approx. Theory $1 (1995),
347-367.
[3] H. Ehlich and K. Zeller, Auswertung der Normen von Interpolationsoperatoren,
Math. Ann. 164 (1966), 105-112.
[4] P. Erd6s and G. Grinwald, Note on an elementary problem of interpolation, Bull.
Amer. Math. Soc. 44 (1938), 515-518.
[5] G. Faber, ber die interpolatorische Darstellung stetiger Funktionen, Jahresber.
Deutsch. Math.-l/’erein. 23 (1914), 190-210.
[6] R. Giinttner, Evaluation of Lebesgue constants, SlAM J. Numer. Anal. 17 (1980),
512-520.
[7] R. Krel3, On general Hermite trigonometric interpolation, Numer. Math. 20 (1972),
125-138.
[8] M. Riesz, Eine trigonometrische Interpolationsformel und einige Ungleichungen fiir
Polynome, Jahresber. Deutsch. Math.-l/’erein. 23 (1914), 354-368.
[9] T.J. Rivlin, An Introduction to the Approximation of Functions, Dover, New York,
1981.
[10] T.J. Rivlin, Chebyshev Polynomials, 2nd ed., Wiley, New York, 1990.
[11] S.J. Smith, On the positivity of the fundamental polynomials for generalized
Hermite-Fejr interpolation on the Chebyshev nodes, J. Approx. Theory (to appear).
[12] J. Szabados, On the order of magnitude of fundamental polynomials of Hermite
interpolation, Acta Math. Hungar. 61 (1993), 357-368.